Is it possible to express $\mathcal{L} f(t+a)$ using $F$ (Laplace transforms)

Question

Typical notation for Laplace transforms is that $\mathcal{L} f(t) = F(s)$

Is it possible to rewrite $\mathcal{L} f(t+a)$, where $a$ is some arbitrary constant, to have an expression involving $F$? Clearly it would not just be $F(s+a)$... I think

Answer 1

$\mathcal L(f(t)) = \int_0^\infty f(t) e^{-st} \ dt = F(s)$. So

\begin{align*} \mathcal L(f(t+a)) &= \int_0^\infty f(t+a) e^{-st} \ dt \\ & = \int_a^\infty f(p) e^{-s(p-a)} \ dp \\ & = e^{as} \left[ F(s) - \int_0^a f(p) e^{-sp} \ dp \right] \end{align*}

EDIT: Sorry, the last integration switch was a typo. The limits of integration were changed via the substitution $p = t + a$ in the second step. In the final step, we used the fact that $$\int_0^\infty = \int_0^a + \int_a^\infty.$$